Question

The derivative of $\cos 3x$ with respect to $\sin 3x$ is

Answer 1

Step 1. Let $u=\cos 3xandv=\sin 3x$

Step 2. Differentiating $u=\cos 3x$with respect to $x$ we have

$$\begin{gathered} \frac{du}{dx}=\frac{d}{dx}\cos 3x \\ \Rightarrow \frac{du}{dx}=-3\sin 3x...................\left(i\right) \end{gathered}$$

Step 3. Differentiating $v=\sin 3x$with respect to $x$ we have

$$\begin{gathered} \frac{dv}{dx}=\frac{d}{dx}\sin 3x \\ \Rightarrow \frac{dv}{dx}=3\cos 3x...................\left(i\right) \end{gathered}$$

Step 4.Dividing $\left(i\right)$ by $\left(i\right)$ we have

$$\begin{gathered} \frac{du}{dx}x\frac{dx}{dv}=\frac{-3\sin 3x}{3\cos 3x} \\ \Rightarrow \frac{du}{dv}=\frac{-\sin 3x}{\cos 3x} \\ \Rightarrow \frac{du}{dv}=-\tan 3x \end{gathered}$$

Hence The derivative of $\cos 3x$ with respect to $\sin 3x$ is $-\tan 3x$